I drew the attachment, but I don't know how to proceed form there. I don't understand how to choose the region U<=u or where u splits.
I know for u<0, F_U(u)=0 since Y_1-Y_2<0 has no solution and for u>2, F_U(u)=1

Re: Method of Distribution Functions problem

Well, 0<Y1-Y2<2 so 0<u<2
I guess that's the region for U and if I let Y2=0 I get the intercept u=1, so I guess that's where it splits.
So for 0<u<1 the region is above the line Y1=u+Y2 and for 1<u<2 the region is below that line. Is that right?

Re: Method of Distribution Functions problem

Now you can integrate over that region with respect to a varying value of u (with your limits) to get a CDF and differentiate to get a PDF in terms of u.

Can you write an integral in terms of u given the limits of joint distribution that satisfy Y1 - Y2 being in the set?

(Hint: Think about each value of u as a slice for each valid value of Y1 - Y2 in the integration region and the probability for a particular u sums up all the probabilities in that slice. To sum something up, you integrate the probability for that slice and that becomes your PDF for P(U = u)).

Re: Method of Distribution Functions problem

Ok, so the drawing was bad and it was throwing me off so I attached a better one. So the area Y_1-Y_2<u is always above the line Y_1-Y_2=u and with Y_2=0 we have that u=Y_1. After the line sort of passes the point (2,1) though, the integration limits will not be the same and at that point u=1, so that's why it has to split there.

a) So for 0<u<1 we have the limits 0<y_2<u and 2y_2<y_1<u+y_2 for 1dy_1dy_2. Integrating part is easy so F_U(u)=(u^2)/2

b) and for 1<u<2 the upper area gets complicated to integrate so you take the complement (A in the attachment) the limits are y_2+u<y_1<2 and 0<y_2<2-u again for 1dy_1dy_2. So F_U(u)=1-((2-u)^2)/2, the one minus is from the fact that it is the compliment.

You differentiate those and get the pdf for u.

Thanks chiro for helping me out. I thought I would share this since it took me forever to figure it out.